Question

Решите уравнение f'(x)=0, если f(x)=cos^3x-sin^3x

Answer 1

F'(x)=3cos^2(x)*(-sin(x))-3sin^2(x)*cos(x)= -3(cos^2(x)sinx+sin^2(x)cosx)
-3(cos^2(x)sinx+sin^2(x)cosx)=0
cos^2(x)sinx+sin^2(x)cosx=0
sinxcosx(cosx+sinx)=0

sinxcosx=0
sinx=0
x=pi*n, n∈Z

cosx=0
x=pi/2+pi*k, k∈Z

sinx= -cosx |:cosx≠0
tgx= -1
x= -pi/4+pi*t, t∈Z

Answer 2

To find the critical points of the function $f(x) = \cos^3(x) - \sin^3(x)$, we need to find the values of $x$ where the derivative of $f(x)$ is equal to zero ($f'(x) = 0$).

First, let's find the derivative $f'(x)$: $f(x) = \cos^3(x) - \sin^3(x)$ Using the chain rule, the derivative is: $f'(x) = 3\cos^2(x)(-\sin(x)) - 3\sin^2(x)(\cos(x))$ $f'(x) = -3\cos^2(x)\sin(x) + 3\sin^2(x)\cos(x)$

Next, we set $f'(x) = 0$ and solve for $x$: $-3\cos^2(x)\sin(x) + 3\sin^2(x)\cos(x) = 0$ $3\cos(x)\sin(x)(\sin(x) - \cos(x)) = 0$

Now, there are two possibilities for this equation to be satisfied:

1. $3\cos(x)\sin(x) = 0$
2. $\sin(x) - \cos(x) = 0$

Let's solve each part:

1. $3\cos(x)\sin(x) = 0$

This equation has two solutions: a) $\cos(x) = 0$, which gives $x = \frac{\pi}{2} + k\pi$, where $k$ is an integer. b) $\sin(x) = 0$, which gives $x = k\pi$, where $k$ is an integer.

2. $\sin(x) - \cos(x) = 0$

This equation has one solution: $\sin(x) = \cos(x)$

Divide both sides by $\cos(x)$: $\tan(x) = 1$

The solutions for $\tan(x) = 1$ are $x = \frac{\pi}{4} + k\pi$, where $k$ is an integer.

So, the critical points of $f(x) = \cos^3(x) - \sin^3(x)$ are: $x = \frac{\pi}{2} + k\pi, \quad x = k\pi, \quad x = \frac{\pi}{4} + k\pi$ where $k$ is an integer.